Problem: Simplify; express your answer in exponential form. Assume $a\neq 0, q\neq 0$. $\dfrac{{(a^{2})^{-2}}}{{(a^{2}q^{-3})^{2}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${a^{2}}$ to the exponent ${-2}$ . Now ${2 \times -2 = -4}$ , so ${(a^{2})^{-2} = a^{-4}}$ In the denominator, we can use the distributive property of exponents. ${(a^{2}q^{-3})^{2} = (a^{2})^{2}(q^{-3})^{2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(a^{2})^{-2}}}{{(a^{2}q^{-3})^{2}}} = \dfrac{{a^{-4}}}{{a^{4}q^{-6}}}$ Break up the equation by variable and simplify. $\dfrac{{a^{-4}}}{{a^{4}q^{-6}}} = \dfrac{{a^{-4}}}{{a^{4}}} \cdot \dfrac{{1}}{{q^{-6}}} = a^{{-4} - {4}} \cdot q^{- {(-6)}} = a^{-8}q^{6}$.